Answer:
the second one
Explanation:
im positive bc ur gaining 2 ions
I have provided two images to help with this question. The first image is the reaction that is taking place. The γ-pyran is treated with the hydride acceptor triphenylmethyl perchlorate. A hydride is a hydrogen atom containing a lone pair of electrons giving it a negative charge. The triphenylmethyl cation is a positively charged carbocation that greatly wants to accept an electron pair to stabilize its charge. Therefore, it abstracts a hydride from the γ-puran starting material. It grabs one of the hydrogen atoms that is drawn in the reaction scheme. This results in the formation of triphenylmethane and a pyrylium perchlorate salt with the formula C₅H₅ClO₅. The important aspect of the structure is shown in the attached images. The most stable resonance form of the pyrylium cation is shown with a positive charge on the oxygen.
The reason this pyrylium ion is the most stable resonance form is because the formation of the oxonium ion (positive charged oxygen with 3 bonds) leads to an aromatic compound. There are 6 pi electrons in conjugation in this ring similar to a benzene ring and this results in the most stable structure.
Reaction for precipitation is
Pb(NO3)2(aq) + 2Nacl(aq)→PbCL2(s)+2NaNO3(aq)
The given reaction is
The mol of Nacl required is =2×mol of Pb(NO3)2 which is present.
∴M(Nacl)×V(Nacl) =2×M(Pb(NO3)2 ×V(Pb(NO3)2
o.100M×V(Nacl)=2×0.200 MX75.0ML
V(Nacl = 300mL.
Answer:
Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)
Explanation:
<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.
Calculating Percent Composition of NaHCO₃:
1: Calculating Molar Masses of all elements present in NaHCO₃:
a) Na = 22.99 g/mol
b) H = 1.01 g/mol
c) C = 12.01 g/mol
d) O₃ = 16.0 × 3 = 48 g/mol
2: Calculating Molecular Mass of NaHCO₃:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O₃ = 48 g/mol
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Total 84.01 g/mol
3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:
For Na:
= 22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 27.36 %
For H:
= 1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 1.20 %
For C:
= 12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 14.29 % ≈ 14.30 %
For O:
= 48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 57.13 % ≈ 57.14 %